Answer:
The net torque is zero
Explanation:
Let's assume that the dipole is compose of two equal but oposite charges e, and it cam be represented by a rod with one end having a charge e and the other end with a charge of -e. Notice that the dipole is parallel to the electric field thus the force felt by both of the charges will be parallel to the electric field. This means that there will be no components of the forces that are perpendicular to the rod which is a requirement for it to have a torque.
Answer:
Explanation:
Since the wires attract each other , the direction of current will be same in both the wires .
Let I be current in wire which is along x - axis
force of attraction per unit length between the two current carrying wire is given by
x 
where I₁ and I₂ are currents in the wires and d is distance between the two
Putting the given values
285 x 10⁻⁶ = 10⁻⁷ x 
I₂ = 16.76 A
Current in the wire along x axis is 16.76 A
To find point where magnetic field is zero due the these wires
The point will lie between the two wires as current is in the same direction.
Let at y = y , the neutral point lies
k 2 x 
= k 2 x 
25.5y = 16.76 x .3 - 16.76y
42.26 y = 5.028
y = .119
= .12 m
Answer:
The mass of a single paper is approximately 0.047 lb/paper which in SI Units is approximately 21.77 g/paper
Explanation:
The given information on the size and the weight of paper are;
The mass of a box of 500 sheets of paper = 24 lb
The number of sheets in the paper = 500 sheets
The dimensions of the paper = 17 in. × 22 in., which is equivalent to 43.18 cm × 55.88 cm
The mass of a single paper = The mass of the box of paper/(The number of sheets of paper present in the box)
The mass of a single paper = 24 lb/500 = 0.047 lb/paper
Given that 1 lb = 453.6 g, we have;
0.047 lb/paper = 0.047 lb/paper×453.6 g/(lb) = 21.77 g/paper
The mass of a single paper = 0.047 lb/paper = 21.77 g/paper.
Thermal energy that flows between objects due to a difference in temperature is heat.
Yes since the heavy truck it’s way bigger it can cause more damage to the passenger car
